Topic 3: Interdependent values

Topic 3: Interdependent values • Introductory stuff: • a newspaper article • experiment of the winner’s curse • Definition of model • Analysis of revenue for the classic auctions

הסיפור מאחורי הפרטת בזק • מקור: "הארץ", 29 למאי 2005 • שתי קבוצות התמודדו במכרז: קבוצת סבן, וקבוצת אלג'ם. • ההצעות ההתחלתיות - אלג'ם: 3.2 מיליארד , סבן: 3.6 מיליארד • התכנון המקורי של משרד האוצר היה לחשוף את ההצעות, ולקבל הצעות חדשות (כלומר לערוך מכרז אנגלי עולה). אבל לנוכח הפער הגדול בין שתי ההצעות (400 מיליון שקל), באוצר חששו להמשיך לפי התכנון המקורי, שמא אלג'ם תתייאש ותפרוש. • היועץ המקצועי, פרופ' מוטי פרי מירושלים, התעקש להמשיך לפי התכנון: "אין ביטחון מלא, אבל בדרך כלל השקיפות מביאה לתוצאות הטובות ביותר. יש למסור לכל אחד מהמשתתפים את תוצאות הסבב הראשון... ההצעה העיקרית שלי במכרז היתה ליצור מערכת שקופה לחלוטין, שבה כל אחד מהצדדים יודע בדיוק מה קורה ומקבל את כל המידע. ככל שלמתחרים יש יותר ידע, רמת הסיכון שלהם יורדת, מרכיב ההימור בעסקה יורד, והם מוכנים להסתכן בהצעות גבוהות יותר".

הסיפור מאחורי הפרטת בזק • ההצעות ההתחלתיות - אלג'ם: 3.2 מיליארד , סבן: 3.6 מיליארד. • המשך המכרז: אלג'ם העלו ל-3.8 מיליארד, וסבן העלה ל-4.11 מיליארד, וניצח. • מכרז בזק היה המכרז הראשון בתולדות ההפרטה שנוהל על פי ייעוץ של פרופ' לתורת המשחקים. תוצאות המכרז הוכיחו שגישה זו הייתה נכונה. • למרות כל זה, יש לשים לב שהמודל הנכון למכרז הזה הוא לא בדיוק המודל שלנו, כי זה לא נכון שערך של בזק לשחקנים מוגרל בצורה בלתי תלויה מהתפלגות כלשהי. יותר נכון להגיד שיש ערך מסוים לבזק, וכל אחד מהשחקנים מבצע אומדן לערך זה.

Interdependent symmetric valuations • Vi(X1,…,Xn)=V(Xi , X-i) • V is symmetric in the last N-1 components. • V(0,…,0)=0. • V is non-decreasing. • Example: Vi(X1,…,Xn) = (i Xi )/n • This is more or less what we saw in class earlier. Each student had an estimate to the sum of money in the jar, and the correct amount was most probably the average of all estimates.

The winner’s curse • Start with an example. Suppose two players, the value function is Vi(X1,X2) = (X1+X2)/2, and the signals satisfy X1 < X2. Suppose the first player knows the second player will drop when the price reaches X2. When should he drop?

The winner’s curse • Start with an example. Suppose two players, the value function is Vi(X1,X2) = (X1+X2)/2, and the signals satisfy X1 < X2. Suppose the first player knows the second player will drop when the price reaches X2. When should he drop? • If he will drop after the second player, then he will pay more than the value of the object.

The winner’s curse • Start with an example. Suppose two players, the value function is Vi(X1,X2) = (X1+X2)/2, and the signals satisfy X1 < X2. Suppose the first player knows the second player will drop when the price reaches X2. When should he drop? • If he will drop after the second player, then he will pay more than the value of the object. • Notice that X2 can be very close to X1, so if player does not want to pay more than the worth of the object, he has to drop when the price reaches his signal.

The winner’s curse • Start with an example. Suppose two players, the value function is Vi(X1,X2) = (X1+X2)/2, and the signals satisfy X1 < X2. Suppose the first player knows the second player will drop when the price reaches X2. When should he drop? • If he will drop after the second player, then he will pay more than the value of the object. • Notice that X2 can be very close to X1, so if player does not want to pay more than the worth of the object, he has to drop when the price reaches his signal. • However, in that point, since player 2 did not drop yet, then player 1 knows that the value is larger than the price. Therefore he is tempted to stay more in auction. If he indeed stays, and player 2 drops, he ends up paying more than his value. This is called the “winner’s curse”.

The English auction • Suppose we have 3 players, Vi = 1/3(X1 + X2 + X3). • Consider the following strategies: • In the beginning, each player will drop at her signal. • After the first player drops, the other two can infer her signal, and so they can update their value. As a result, each i of the remaining two will drop at 1/3(X1 + Xi + Xi). • With signals 10, 20, and 30, we will have: the first player will drop at price=10, the second player will drop at price=50/3, and the first player will win and will pay this. Notice that her actual value is 20 > 50/3.

The English Auction • Let bN(x) = V(x,…,x). Suppose each bidder j ≠ N follows a strategy were he drops out exactly when the price reaches bN(Xj). • Player N stays in the auction when the price is p if and only if he has positive utility for winning with price p. Choose y s.t. p=bN(y). • Winning means that all players have dropped, so their signals are at most y. Therefore N’s value is at most V(XN,y,…,y). • Thus player N will stay in the auction iff p <V(XN,y,…,y), => y < XN, meaning that he will quit at pN = V(XN,…, XN)

The English Auction • Let bN(x) = V(x,…,x). Suppose each bidder j ≠ N follows a strategy were he drops out exactly when the price reaches bN(Xj). • Player N stays in the auction when the price is p if and only if he has positive utility for winning with price p. Choose y s.t. p=bN(y). • Winning means that all players have dropped, so their signals are at most y. Therefore N’s value is at most V(XN,y,…,y). • Thus player N will stay in the auction iff p <V(XN,y,…,y), => y < XN, meaning that he will quit at pN = V(XN,…, XN) • Conclusion: the player withthe lowest signaldrops first (suppose it is N). Now, The others see pN and therefore infer XN.

The English Auction • Let bN(x) = V(x,…,x). Suppose each bidder j ≠ N follows a strategy were he drops out exactly when the price reaches bN(Xj). • Player N stays in the auction when the price is p if and only if he has positive utility for winning with price p. Choose y s.t. p=bN(y). • Winning means that all players have dropped, so their signals are at most y. Therefore N’s value is at most V(XN,y,…,y). • Thus player N will stay in the auction iff p <V(XN,y,…,y), => y < XN, meaning that he will quit at pN = V(XN,…, XN) • Conclusion: the player withthe lowest signaldrops first (suppose it is N). Now, The others see pN and therefore infer XN. • Let bN-1(x, pN) = V(x,…,x, XN). By the same logic, the player with the second lowest signal (say N-1) drops at bN-1(XN-1, pN), and so on…

Example • Four players with signals: X1 = 10, X2 = 8, X3 = 4, X4 = 2. • True value is the average of the signals (in this case, it is 6). • How will the strategy look like?

Example • Four players with signals: X1 = 10, X2 = 8, X3 = 4, X4 = 2. • True value is the average of the signals (in this case, it is 6). • How will the strategy look like? • Step 1: each player decides to drop at her signal. Player 4 drops first when the price is 2. The others observe that, and infer that her signal was 2.

Example • Four players with signals: X1 = 10, X2 = 8, X3 = 4, X4 = 2. • True value is the average of the signals (in this case, it is 6). • How will the strategy look like? • Step 1: each player decides to drop at her signal. Player 4 drops first when the price is 2. The others observe that, and infer that her signal was 2. • Step 2: player i=1,2,3 decides to drop when the price will be(3·Xi + 2)/4. Player 3 will drop second, when the price is 3.5. The other players infer that her signal is 4.

Example • Four players with signals: X1 = 10, X2 = 8, X3 = 4, X4 = 2. • True value is the average of the signals (in this case, it is 6). • How will the strategy look like? • Step 1: each player decides to drop at her signal. Player 4 drops first when the price is 2. The others observe that, and infer that her signal was 2. • Step 2: player i=1,2,3 decides to drop when the price will be(3·Xi + 2)/4. Player 3 will drop second, when the price is 3.5. The other players infer that her signal is 4. • Step 3: player i=1,2 decides to drop when the price will be(2·Xi + 4 + 2)/4. Player 2 will drop third, when the price is 5.5. Now player 1 is the only one that remains, and he wins. His profit is 0.5.

Analysis • In general, suppose players drop at prices pN, pN-1,…,p2.Define: bk(x , pk+1,…, pN) = V(x,…,x, xk+1 ,…, xN)and let xksolve: bk(xk , pk+1,…, pN) = pk • Let Yi = maxj≠iXj. If all players follow the strategys* = [b1(),…,bN()] then the player with the highest signalwins (suppose it is player i), and his price is p2 = V(Yi, X-i). THM The strategy s* forms a symmetric equilibrium.

Analysis • In general, suppose players drop at prices pN, pN-1,…,p2.Define: bk(x , pk+1,…, pN) = V(x,…,x, xk+1 ,…, xN)and let xksolve: bk(xk , pk+1,…, pN) = pk • Let Yi = maxj≠iXj. If all players follow the strategys* = [b1(),…,bN()] then the player with the highest signalwins (suppose it is player i), and his price is p2 = V(Yi, X-i). THM The strategy s* forms a symmetric equilibrium. Proof: Fix a player i and suppose all the others are playing this. Case I: Xi > Yi. Therefore i wins if playing s*, and V(Xi, X-i) >p2.p2 doesn’t depend on i’s actions so he cannot improve his utility.

Analysis • In general, suppose players drop at prices pN, pN-1,…,p2.Define: bk(x , pk+1,…, pN) = V(x,…,x, xk+1 ,…, xN)and let xksolve: bk(xk , pk+1,…, pN) = pk • Let Yi = maxj≠iXj. If all players follow the strategys* = [b1(),…,bN()] then the player with the highest signalwins (suppose it is player i), and his price is p2 = V(Yi, X-i). THM The strategy s* forms a symmetric equilibrium. Proof: Fix a player i and suppose all the others are playing this. Case I: Xi > Yi. Therefore i wins if playing s*, and V(Xi, X-i) >p2.p2 doesn’t depend on i’s actions so he cannot improve his utility. Case II: Xi< Yi. Therefore if i will play s* he will lose. The only way to improve utility is to win, but winning implies paying V(Yi, X-i) > V(Xi, X-i) so i’s utility will be at most 0.

Remarks • When a player drops he knows that his value is higher than the price! (but still he has no way to win the auction with a profit) • This equilibrium is in fact stronger than Bayesian-Nash:DFN: The strategies s1,…, snare in ex-post equilibrium if for any i, v-i,vi, ai : ui(si(vi),s-i(v-i) > ui(ai,s-i(v-i) • Implies a Bayesian-Nash equilibrium for any possible distribution. • Has the “no regret” property: a player does not regret her action even after knowing the signals and actions of the other players.

Remarks (2) • The English auction does not have dominant-strategies: if the player with the lowest signal stays after he is supposed to retire, the rest will get a false picture of his signal, and can pay more than their value.For example, X1=x, X2=y, x>y, and player 2 drops first at v(y’,y’) such that y<y’<x and v(y,y) < v(x,y) < v(y’,y’) < v(x,x)

Affiliation • Affiliation: We assume the signals are positively correlated. Formally, we will use the following properties: • Let v*(x, y) = E [ Vi | Xi = x and Yi = y ].Then v*(x,y) is non-decreasing in x,y. • For x > y: v*(y,y) = E[V(Xi , X-i) | Xi = y and Yi = y ] = E[V(Yi , X-i) | Xi = y and Yi = y ]< E[V(Yi , X-i) | Xi = x and Yi = y ]

Second-price auction • Reminder: each player places a bid. The highest bidder wins. Pays the second highest price • In private values: a dominant strategy to bid the true value. • Here: what is the true values? The player doesn’t know. • One is tempted to bid E( Vi | Xi = x ). • But then, suppose player i wins. This means that he has the highest value, and so E( Vi | Xi = x ) > E( Vi | Xi = x and Yi < x) • So winning is an indication that the value is not as high as you first thought… This is again the winner’s curse. • By this reasoning, it seems that each player should bid b(x)=v*(x,x). Will this work?

Analysis Intuition: If the highest bid among all other players is b(y), player i infers that her expected value is v*(Xi,y). Hence:If y < Xi then v*(y,y) = b(y) < v*(Xi,y) => positive utilityIf y > Xi then v*(y,y) = b(y) > v*(Xi,y) => negative utility=> winning is good up to y=Xi, so bid v*(Xi, Xi)

Analysis Intuition: If the highest bid among all other players is b(y), player i infers that her expected value is v*(Xi,y). Hence:If y < Xi then v*(y,y) = b(y) < v*(Xi,y) => positive utilityIf y > Xi then v*(y,y) = b(y) > v*(Xi,y) => negative utility=> winning is good up to y=Xi, so bid v*(Xi, Xi) THM: Symmetric equilibrium in a second-price auction is given by: b(x)=v*(x,x)

Analysis Intuition: If the highest bid among all other players is b(y), player i infers that her expected value is v*(Xi,y). Hence:If y < Xi then v*(y,y) = b(y) < v*(Xi,y) => positive utilityIf y > Xi then v*(y,y) = b(y) > v*(Xi,y) => negative utility=> winning is good up to y=Xi, so bid v*(Xi, Xi) THM: Symmetric equilibrium in a second-price auction is given by: b(x)=v*(x,x) Proof: If player i bids b(z), and all others follow their equilibrium strategies, she wins if b(Yi) < b(z)  Yi< z. => i’s expected utility when she bids b(z) and her signal is Xiis 0z[v*(Xi,y) - v*(y,y)]·g(y| Xi)·dy and this is maximized for z= Xi, by the argument above.

Remarks • In contrast to private values, here the second-price auction only has a Bayesian-Nash Equilibrium, and no dominant-strategies. • Players need to know the probability distributions of the signals in order to find v*. • In private values, V(Xi , X-i)=Xi , and v*(x, y) = x. So this generalizes the result for private values.

Revenue comparison THM: The expected revenue of the English auction is at least that of the second-price auction. Proof: We have: E[RII | i wins] = E[ v*(Yi, Yi) | Xi > Yi ] E[ REnglish | i wins] = E[ V(Yi , X-i) | Xi> Yi ] by affiliation, for x>y: v*(y,y) < E[V(Yi , X-i) | Xi=x and Yi=y ] => E[ v*(Yi, Yi) | Xi > Yi ] < E[ V(Yi , X-i) | Xi> Yi ].

First-price auctions • First-price auctions have in equilibrium a revenue which is even smaller than the second-price auction. • Proof is complicated and long, and I will not teach it. • Thus among the classic auctions, the English auction is best.

Information Revelation • The equilibrium of the English auction is equal to the following situation: nature reveals the N-2 lowest signals to everyone, and then we conduct a second-price auction among the two bidders with the two highest signals. • The previous theorem can be stated as saying: information revelation increases the revenue for interdependent values. • This is true also, for example, if the auctioneer has some additional information about the item, X0.

Efficiency • Efficiency: allocating the item to the player with highest value • Measures the society’s welfare, not for the auctioneer’s own utility. • Why? For example, • In the FCC auction, the US law requires the government to maximize the efficiency, and not the revenue. • Super-huge firms sometime have “dummy money” to make inside decisions more efficient (e.g. IBM has “blue-money”) • Ideological reasons: economists should know how to improve global social welfare.

The English auction may be inefficient • With private values, we know that first-price, second-price, and English auction are all efficient. • With general interdependent values (not symmetric), this may not be the case. For example: v1(x1,x2,x3)=x1 + 2·x2x3 + 0.01(x2+x3) v2(x1,x2,x3)=0.5·x1 + x2 v3(x1,x2,x3)=x3 Suppose by contradiction that there exists an efficient equilibrium, and let bi() (i=1,2,3) denote the equilibrium strategies when none of the players have dropped (yet).

v1=x1 + 2·x2x3 + 0.01(x2+x3) ; v2=0.5·x1 + x2 ; v3=x3 Claim 1: Bidder 1 never drops out first. Proof: Suppose x2 , x3 > 0.5 . Then v1 > max(x1 + x2 , x1 + x3 ) > v2 , v3 Thus if 1 drops before knowing x2 , x3 the auction is inefficient.

v1=x1 + 2·x2x3 + 0.01(x2+x3) ; v2=0.5·x1 + x2 ; v3=x3 Claim 1: Bidder 1 never drops out first. Proof: Suppose x2 , x3 > 0.5 . Then v1 > max(x1 + x2 , x1 + x3 ) > v2 , v3 Thus if 1 drops before knowing x2 , x3 the auction is inefficient. Thus for any x1,x2,x3 , b1(x1) > b2(x2),b3(x3). Now suppose x1= 1/8, x2 = 1/4 , x3 = 5/16. • Note that v2 = v3 > v1.

v1=x1 + 2·x2x3 + 0.01(x2+x3) ; v2=0.5·x1 + x2 ; v3=x3 Claim 1: Bidder 1 never drops out first. Proof: Suppose x2 , x3 > 0.5 . Then v1 > max(x1 + x2 , x1 + x3 ) > v2 , v3 Thus if 1 drops before knowing x2 , x3 the auction is inefficient. Thus for any x1,x2,x3 , b1(x1) > b2(x2),b3(x3). Now suppose x1= 1/8, x2 = 1/4 , x3 = 5/16. • Note that v2 = v3 > v1. • It cannot be that b2(x2)=b3(x3) since then player 1 will win.

v1=x1 + 2·x2x3 + 0.01(x2+x3) ; v2=0.5·x1 + x2 ; v3=x3 Claim 1: Bidder 1 never drops out first. Proof: Suppose x2 , x3 > 0.5 . Then v1 > max(x1 + x2 , x1 + x3 ) > v2 , v3 Thus if 1 drops before knowing x2 , x3 the auction is inefficient. Thus for any x1,x2,x3 , b1(x1) > b2(x2),b3(x3). Now suppose x1= 1/8, x2 = 1/4 , x3 = 5/16. • Note that v2 = v3 > v1. • It cannot be that b2(x2)=b3(x3) since then player 1 will win. • It cannot be that b2(x2)<b3(x3) since for a higher x1player 2 has the highest value.

v1=x1 + 2·x2x3 + 0.01(x2+x3) ; v2=0.5·x1 + x2 ; v3=x3 Claim 1: Bidder 1 never drops out first. Proof: Suppose x2 , x3 > 0.5 . Then v1 > max(x1 + x2 , x1 + x3 ) > v2 , v3 Thus if 1 drops before knowing x2 , x3 the auction is inefficient. Thus for any x1,x2,x3 , b1(x1) > b2(x2),b3(x3). Now suppose x1= 1/8, x2 = 1/4 , x3 = 5/16. • Note that v2 = v3 > v1. • It cannot be that b2(x2)=b3(x3) since then player 1 will win. • It cannot be that b2(x2)<b3(x3) since for a higher x1player 2 has the highest value. • It cannot be that b2(x2)>b3(x3) since for a smaller x1player 3 has the highest value.

Check point • Studied interdependent values. • We’ve analyzed the second-price and the English auction. • Second price no longer has dominant strategies. • Second price and English auctions are no longer equivalent. • English auction has an ex-post equilibrium. • English auction obtains in expectation higher revenue than second price. • On the way, we defined ex-post equilibrium, and saw that it is stronger than Bayesian-Nash, and weaker than dominant-strategies • Talked about the efficiency properties of the English auction • Efficiency is a measure for the society, not for the seller • The English auction with private values is efficient, but not with interdependent values.

“optimal” auctions • After analyzing the classic auction formats, we wish to design optimal auctions: • First goal: achieving full efficiency with general interdependent valuations. • Second goal: the possibility of full surplus extraction.

The single-crossing property • We look for a condition that will enable us the design of an efficient mechanism. • The valuation functions satisfy the single-crossing property if: ∂vj/∂xj > ∂vi/∂xj for any i,j and any combination of signals X such that vi(X)=vj(X) and these values are maximal among all values. • In words, j’s signal influences j’s value more than any other signal (at the crucial points).

values vi(x-i) vj(x-i) xi

Remark • The English auction with symmetric valuations that satisfy single-crossing is efficient. • But with general valuations, this is not the case (the valuations in the example we had last week satisfy single-crossing, check at home). • We next construct an efficient mechanism.

An efficient mechanism • Assumption: the mechanism designer knows the valuation functions (but not the signals, of-course). • We construct a direct-revelation mechanism: each player reports her signal. • The winner: the player with the highest value (not the signal). • The price: define the “threshold value” Ti(x-i) = inf { zi | vi(zi, x-i) > max j≠i vj(zi, x-i) }

values vi(x-i) vj(x-i) xi Ti(x-i)

An efficient mechanism(“generalized VCG”) • Assumption: the mechanism designer knows the valuation functions (but not the signals, of-course). • We construct a direct-revelation mechanism: each player reports her signal. • The winner: the player with the highest value (not the signal). • The price: define the “threshold value” Ti(x-i) = inf { zi | vi(zi, x-i) > max j≠i vj(zi, x-i) } and set the price of every player i to be vi( Ti(x-i) , x-i ) if player i wins Mi* (x-i) = 0 if player i loses

values vi(x-i) vj(x-i) Mi* (x-i) xi Ti(x-i)

Analysis THM: Truthfulness is an ex-post efficient equilibrium. Proof: • Efficiency: if all players are truthful then the player with the highest value is the winner. • Truthfulness: The player’s price for winning is independent of her declaration of signal. Assuming that all others are truthful: • If her value is above this price, winning will maximize her utility, and truthfulness will cause her to win. • If her value is below this price, losing will maximize her utility, and truthfulness will cause her to lose.

Remarks • Players are differentiated, and the mechanism may treat each one differently. • No need to know distributions, but we do need to know the exact valuation functions. • This equilibrium is not in dominant-strategies: if a player declares a false signal which is higher than her true one, the winner might pay a price higher than her value. • For private values, this reduces to the second-price auction.

And if single-crossing is violated? Claim: If single-crossing does not hold then there may be no efficient mechanism. Proof: By example: • v1(x1,x2)=x1 ; v2(x1,x2)=(x1)2 ; x1 lies in [0,2] and x2 is a constant. • single-crossing is violated at (1, x2): v1 = v2 = 1 but v1/x1 = 1 < 2 = v1/x2.

And if single-crossing is violated? Claim: If single-crossing does not hold then there may be no efficient mechanism. Proof: By example: • v1(x1,x2)=x1 ; v2(x1,x2)=(x1)2 ; x1 lies in [0,2] and x2 is a constant. • single-crossing is violated at (1, x2): v1 = v2 = 1 but v1/x1 = 1 < 2 = v1/x2. Suppose an efficient mechanism with a price function M1 for player 1: • Fix two signals of player 1, y1 and z1, such that y1 < 1 < z1. • If x1 = z1 then 0 - M1(z1) >z1 - M1(y1) • If x1 = z1 then y1 - M1(y1) > 0 - M1(z1) => y1>z1 , a contradiction. Follows from efficiency and truthfulness

Topic 3: Interdependent values